package 暴力枚举;

import java.util.ArrayList;
import java.util.List;

/**
 * @PackageName: 暴力枚举
 * @ClassName: LeetCode1408
 * @Author: chen jinxu
 * @Date: 2022/8/7 22:45
 * @Description: //数组中的字符串匹配
 * 给你一个字符串数组 words ，数组中的每个字符串都可以看作是一个单词。请你按 任意 顺序返回 words 中是其他单词的子字符串的所有单词。
 * 如果你可以删除 words[j]最左侧和/或最右侧的若干字符得到 word[i] ，那么字符串 words[i] 就是 words[j] 的一个子字符串。
 *      输入：words = ["mass","as","hero","superhero"]
 * 输出：["as","hero"]
 * 解释："as" 是 "mass" 的子字符串，"hero" 是 "superhero" 的子字符串。
 * ["hero","as"] 也是有效的答案。
 *  输入：words = ["leetcode","et","code"]
 * 输出：["et","code"]
 * 解释："et" 和 "code" 都是 "leetcode" 的子字符串。
 *  输入：words = ["blue","green","bu"]
 * 输出：[]
 * 1 <= words.length <= 100
 * 1 <= words[i].length <= 30
 * words[i] 仅包含小写英文字母。
 * 题目数据 保证 每个 words[i] 都是独一无二的。
 */
public class LeetCode1408 {
    public static void main(String[] args) {
        System.out.println(stringMatching(new String[] {"mass","as","hero","superhero"}));
        System.out.println(stringMatching(new String[] {"leetcode","et","code"}));
        System.out.println(stringMatching(new String[] {"blue","green","bu"}));
    }

    public static List<String> stringMatching(String[] words) {
        List<String> result = new ArrayList<>();
        for (int i = 0; i < words.length; i++) {
            for (int j = 0; j < words.length; j++) {
                if (i == j || result.contains(words[i])) {
                    continue;
                }
                if (words[j].indexOf(words[i]) >= 0) {
                    result.add(words[i]);
                    break;
                }
            }
        }
        return result;
    }

}
